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LDRs and resistance

ridgey1988

i thought that the resistance of an LDR lowered as light intensity increased...and vice versa....at least..thats what wikipedia thinks and i have written down in my physics book.
but after gettting my recalled paper back for physics in action.. i have a big X next right over the word "decreases" and another one over "increases" in their respective positions. which is right?

Reply 1

jazznaz

The resistance decreases as light intensity increases. :smile:

Reply 2

Mohit_C

For most semi-conductors (not sure if LDRs are semi-conductors) but they have very few free electrons in millions. Therefore, when you heat up a semi-conductor, if frees up a few electrons, but as these electrons have just been freed, and aren't heavily loaded with energy, heating a semi-conductor increases conductivity, but the resistance decreases as there only are a few moving electrons with little energy/charge. I'm not sure if this is the case in an LDR. Can someone please confirm?

Reply 3

Singh_87

I thought that is true....as light intensity increases, the resistance decreases...or so I have been taught!

Reply 4

silent ninja

What was the question? Did it ask about resistance or current/voltage? Maybe you misread it?

Reply 5

Mohit_C

But are LDRs semi-conductors?

Reply 6

silent ninja

Light Dependent Resistors-- yes they are. Photons that hit the LDR release electrons, thereby increasing the charge carrier density enormously and increasing current.

Reply 7

Mohit_C

Ah right, that makes sense, coz photons carry energy and they hit electrons giving energy to them/giving them Kinetic energy so they cause them to move, so creating a current.

Reply 8

Eau

Agreed:

LDR has the same effect as thermistor. Light and heat are both forms of energy. A burst of energy releases electrons in this "special material". Hence increase in energy --> more delocalised electrons --> increased current --> decreased resistance

Reply 9

silent ninja

Mohit_C

Ah right, that makes sense, coz photons carry energy and they hit electrons giving energy to them/giving them Kinetic energy so they cause them to move, so creating a current.

Yup but i'm not sure its kinetic energy. They're given energy to break away and enable them to conduct. You can use I= nAvq-- n=no. charge carriers per unit volume. When this goes up, there are more electrons that can do work.

With thermistors you can get negative and positive temperature coefficient ones, but ridgey1988 was marked down on a LDR?

Reply 10

Mohit_C

Ok, but what happends to the voltage. It should stay the same I presume coz if one increases the other decreaes, so it maintains the same V. This has got me thinking, so I'm not sure.

Reply 11

silent ninja

in that formula I = nAvq
v= the drift velocity of the electrons

the current increases enormously because R goes down (V/R=I), and thats why often semiconductors are fitted with resistors so they dont blow up. The voltage depends on your supply of course.

Reply 12

Mohit_C

Say if it was a 12 volt supply, and now as the current increases, the resistance decreases, so according to V=IR, as one goes up other goes down, the voltage stays the same right?

Reply 13

silent ninja

V is constant, therefore the current depends on the resistance, I= V/R. When R is tiny, I is huge.

Reply 14

phys981

Be careful with that V remains constant thing.

If the LDR or thermistor is in series with another fixed resistance (ie, like a voltage divider curcuit) then

as its resistance increases, its share of the voltage also increases. Similarly for decreasing resistance, V decreases.

Just thought I should point it out.

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